If in this esterification, initially equal moles of alcohol and acid are taken. At equilibrium, unreacted alcohol and acid, on treatment of sodium, produced H₂ which occupied 44.8 L at 0°C and 1 atm.

Question

If in this esterification, initially equal moles of alcohol and acid are taken. At equilibrium, unreacted alcohol and acid, on treatment of sodium, produced H₂ which occupied 44.8 L at 0°C and 1 atm. The percentage esterification of acid and alcohol is

Prashant Kotian · Accepted Answer

Correct answer: D. Let initially 'a' moles of each be taken, and 'x' moles react. Equilibrium unreacted moles = (a-x) for alcohol and (a-x) for acid. Both react with Na to yield 1/2 H₂ per mole: R-OH + Na → R-ONa + 1/2 H₂, and R-COOH + Na → R-COONa + 1/2 H₂. Total moles of H₂ = 1/2(a-x) + 1/2(a-x) = (a-x). 44.8 L of H₂ at STP is 2 moles. Thus, (a-x) = 2. K_c = 4 = x² / (a-x)² = x² / 2² = x² / 4 => x² = 16 => x = 4. Since a - 4 = 2, a = 6 moles. Percentage esterification = (x / a) × 100 = (4 / 6) × 100 = 66.67%. Therefore, correct answer is D.

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